Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 7

Table of contents
- Solution 7

Solution

Since

w = r u

,

w_{t} + w_{r} = r (u_{t} + u_{r} + \frac{u}{r}),

and radial integration satisfies

d x = 4 π r^{2} d r

. Multiplying the identity from part 6 by

4 π

therefore gives

\frac{d}{d t} \int_{R^{3}} ψ [\frac{1}{2} (u_{t} + u_{r} + \frac{u}{r})^{2} + \frac{∣ u ∣ ^{p + 1}}{p + 1}] d x

= - \frac{1}{2} \int_{R^{3}} ψ^{'} (u_{t} + u_{r} + \frac{u}{r})^{2} d x + \frac{1}{p + 1} \int_{R^{3}} ∣ u ∣^{p + 1} (ψ^{'} - (p - 1) \frac{ψ}{r}) d x,

which is the modified Morawetz identity.

Choose the constant weight

ψ = 1

. The first term on the right vanishes and the second is

- \frac{p - 1}{p + 1} \int_{R^{3}} \frac{∣ u ∣ ^{p + 1}}{r}, d x .

The functional on the left is bounded by the conserved energy using the Hardy inequality in Euclidean space. Integrating in time gives another proof of the Morawetz estimate for the defocusing wave equation.

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